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Thanks.

Ok I will do that

To round to one significant figure  6

To round to two significant figures  6.4

Let b and c be constants such that the quadratic -2x^2 + bx + c

has roots and 3 + sqrt(13) and 3 - sqrt(13).

Find the vertex of the graph of the equation y = -2x^2 + bx + c.

This is a concave down parabola

The axis of symmetry is obviously x=3 because this is a halfway between the two roots.

\(-2x^2 + bx + c=0\\ \text{The roots of any quadraic are given by}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\\ so\\ x=\frac{-b}{-4}\pm\frac{\sqrt{b^2+8c}}{-4}\\ x=\frac{b}{4}\pm\frac{\sqrt{b^2+8c}}{-4}\\ \frac{b}{4}=3\qquad so \qquad b=12\\ x=3\pm\frac{\sqrt{144+8c}}{-4}\\ \frac{\sqrt{144+8c}}{-4}=\sqrt{13}\\ \frac{144+8c}{16}=13\\ \frac{144+8c}{16}=13\\ 8c=64\\ c=8\\~\\ \text{So the quadratic is }-2x^2 + 12x + 8\\ f(x)=-2x^2 + 12x + 8\\ f(3)=26 \)

Here is the graph that I used as a check

LaTex

-2x^2 + bx + c=0\\
\text{The roots of any quadraic are given by}\\
x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\\
so\\
x=\frac{-b}{-4}\pm\frac{\sqrt{b^2+8c}}{-4}\\
x=\frac{b}{4}\pm\frac{\sqrt{b^2+8c}}{-4}\\
\frac{b}{4}=3\qquad so \qquad b=12\\
x=3\pm\frac{\sqrt{144+8c}}{-4}\\
\frac{\sqrt{144+8c}}{-4}=\sqrt{13}\\
\frac{144+8c}{16}=13\\
\frac{144+8c}{16}=13\\
8c=64\\
c=8\\~\\
\text{So the quadratic is   }-2x^2 + 12x + 8\\
f(x)=-2x^2 + 12x + 8\\
f(3)=26

Just add LIKE terms.  If the powers of x are not the same then the terms are not LIKE terms.

Triangles ABC and ABD are isosceles with AB = AC = BD, and BD intersects AC at E.  If BD is perpendicular to AC, and angle C = 70 degrees, then find angle EAD.

angle EAD = 25º

Let b and c be constants such that the quadratic -2x^2 + bx + c has roots and 3 + sqrt(13] and 3 - sqrt(13). Find the vertex of the graph of the equation y = -2x^2 + bx + c.

Hello Guest!

\(y=0=(x+3+\sqrt{13})(x-3+\sqrt{13})\\ y=x^2-3x+x\sqrt{13}+3x-9+3\sqrt{13}+x\sqrt{13}-3\sqrt{13}+13\\ y=x^2+x(-3+\sqrt{13}+3+\sqrt{13})+\sqrt{13}(3-3)-9+13\\ \color{blue}y=x^2+2\sqrt{13}x+4\)

\(V(-\frac{b}{2},-(\frac{b}{2})^2+c)\\ V(-\frac{2\sqrt{13}}{2},-(\frac{2\sqrt{13}}{2})^2+4)\\ \color{blue}V(-\sqrt{13},-9)\)

  !

Please post only one question per post.

Let line L1 be the graph of 5x + 8y = -9. Line L2 is perpendicular to line L1 and passes through the point (10,10). If line L2 is the graph of the equation y=mx +b, then find m+b.

Hello Guest!

\(L1\\ 5x + 8y = -9\\ y=-\frac{5}{8}x-\frac{9}{8}\)

\(\color{blue}m=-\frac{1}{-\frac{5}{8}}\\ \color{blue}m=1.6\)

\(L2\\ y=mx +b\\ P(10,10)\\ 0=1.6\cdot 10+b\\ b=10-1.6\cdot 10\)

\(\color{blue}b=-6\\ y=1.6x-6\)

\(m+b=-4.4\)

  !

Is it a triangle?

You need more brackets to make your meaning clear.

A function f has domain [0, 2] and range [0, 1]. (The notation [a, b] denotes {x : a ≤ x ≤ b}.)

Define the function g as g(x) = 1 − f(x + 1).

If [p, q] is the domain of g and [r,s] is the range of g, then determine p + q + r + s.

I am not really sure but,

f(x) has a domain of  [0,2]

so

\(For\;\;f(x)\quad0\le x\le2\\ For\;\;f(x+1)\\ 0\le x+1 \le2\\ -1\le x \le1\)

So the domain of g is [-1,1]      p=-1,  q=1

----

\(0\le f(x+1) \le 1\\ \text{seperates to}\\ 0\le f(x+1) \qquad \qquad f(x+1) \le 1\\ 0\ge -f(x+1) \quad \qquad -f(x+1) \ge -1\\ 1\ge 1-f(x+1) \qquad 1-f(x+1) \ge 0\\ 1-f(x+1)\le 1 \qquad 0\le1-f(x+1)\\ 0\le1-f(x+1)\le 1\\ 0\le g \le 1\qquad \qquad \\\text{The range of g is [0,1]} \)

-1+1+0+1= 1

----------------------

Here is a graph that fits the description to help you understand what I have done

I chose f to be the top half of a circle, radius 1 and centre (1,0)

Here is the direct link to the Desmos graph if you want it.

angle CXY = 70º

By using point-slope form, we get that m + b = -7/3.

By the sine law, AB = 4 + sqrt(3).

Hi!

BUSTED!! this is an AoPS alcumus question

but here:

as volume increases, the ratio becomes greater. Think about what would make the volume very large. It could be large to the point where it could be barely contained in the sphere. You can try to find the approximate ratio now!

The probability is 43/501

Let midpoints of SR and RT be M and N

MAP and NBP are right angles

∠AMS = 63º              ∠BNT = 42º

AMNBP is a pentagon         (sum of interior angles of a pentagon is 540º) 

∠APB = 540 - (180 + 117 +138) = 105º

We need to use casework. Suppose first that bag A is chosen: there is a $1/2$ chance of this occurring. There are ${5 \choose 2} = \frac{5 \cdot 4}{2} = 10$ total ways to select two balls from bag A. If the two balls are the same color, then they must be either both white or both red. If both are white, then there are ${3\choose 2} = 3$ ways to pick the two white balls, and if both are red, then there is $1$ way to pick the two red balls. Thus, the probability of selecting two balls of the same color from bag A is $\frac{1+3}{10} = \frac{2}{5}$.  Next, suppose that bag B is chosen, again with $1/2$ chance. There are ${9 \choose 2} = \frac{9 \cdot 8}{2} = 36$ ways to pick the two balls. There are ${6 \choose 2} = \frac{6 \cdot 5}{2} = 15$ ways to pick two white balls, and ${3 \choose 2} = 3$ ways to pick two red balls. Thus, the probability that two balls drawn from bag B are the same color is equal to $\frac{15+3}{36} = \frac 12$.  Thus, the probability that the balls have the same color is $\frac 12 \cdot \frac 25 + \frac 12 \cdot \frac 12 = \boxed{\frac 9{20}}$.

Hope it helps :)

Here is the link to the answer: https://web2.0calc.com/questions/how-many-ways-are-there-to-put-5-balls-in-3-boxes-if-the-balls-are-not-distinguishable-and-neither-are-the-boxes.

By the cosine law, the largest angle is 135 degrees.

The total number of paths from P to Q is 348.

The total weight of the berries after two days is 70 kg.